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lesya692 [45]
2 years ago
7

I need HELP!!!!!!!!!

Mathematics
1 answer:
erica [24]2 years ago
4 0

Answer:

b

Step-by-step explanation:

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Tomas is finding the coordinates of point A. He knows that point A has an x-coordinate of 8. He starts to compute the value of t
Contact [7]

The complete question is

"Tomas is finding the coordinates of point A. He knows that point A has an x-coordinate of 8. He starts to compute the value of the y-coordinate. Complete Tomas’s work to answer the question.

Which is the coordinate of point A? (8, –6) (–6, 8) (6, –8) (–8, –6)"

<h3>How to find the coordinates?</h3>

The order can be written as, (x,y).

Tomas is trying to find the coordinates of point A.

x- coordinate of Point A is given as 8.

Among the four given options, only option A has x, coordinate as 8.

So, Coordinate of Point A = (8, -6).

Hence, the correct Option is A (8, -6).

Learn more about Coordinate ;

brainly.com/question/7853289

#SPJ1

7 0
2 years ago
Answer ASAP pleaseeee !!!!
Eddi Din [679]

x=8.9 because when you square it and divide you get 39.6.. rounded to 40.

3 0
3 years ago
Read 2 more answers
Which ratio can you get by scaling up the ratio 5/6 / 1/3? Select all that apply.
Liula [17]

Answer:

the answer is 5/2 or alternative form 2*1/2 so the answer is C

3 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP 50 POINTS BRAINLIEST IF CORRECT
beks73 [17]
I think the answer is B
7 0
2 years ago
Read 2 more answers
Can anyone figure this out?
Verizon [17]

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10})\qquad A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ NA=\sqrt{(6+3)^2+(3-10)^2}\implies NA=\sqrt{130} \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1}) \\\\\\ AD=\sqrt{(6-6)^2+(-1-3)^2}\implies AD=4 \\\\[-0.35em] ~\dotfill


\bf D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1})\qquad N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10}) \\\\\\ DN=\sqrt{(-3-6)^2+(10+1)^2}\implies DN=\sqrt{202}


now that we know how long each one is, let's plug those in Heron's Area formula.


\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{130}\\ b=4\\ c=\sqrt{202}\\[1em] s=\frac{\sqrt{130}+4+\sqrt{202}}{2}\\[1em] s\approx 14.81 \end{cases} \\\\\\ A=\sqrt{14.81(14.81-\sqrt{130})(14.81-4)(14.81-\sqrt{202})} \\\\\\ A=\sqrt{324}\implies A=18

5 0
3 years ago
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